求 xtanx(secx)^4的不定积分

用凑微分
∫xtanx(secx)^4dx
=∫xtanx([(tanx)^2+1]dtanx
=∫x(tanx)^3dtanx+∫xtanxdtanx
=(1/4)∫xd(tanx)^4+(1/2)∫xd(tanx)^2
=(1/4)[x(tanx)^4-∫(tanx)^4dx]+(1/2)[x(tanx)^2-∫(tanx)^2dx].(*)
因为
∫(tanx)^4dx=∫(tanx)^2((secx)^2-1)dx=∫(tanx)^2dtanx-∫(tanx)^2dx=(1/3)(tanx)^3-∫(tanx)^2dx
而∫(tanx)^2dx=∫[(secx)^2-1]dx=tanx-x+c1
代入(*)所以原积分=(*)
=(1/4)x(tanx)^4-(1/4)∫(tanx)^4dx+(1/2)x(tanx)^2-(1/2)∫(tanx)^2dx]
=(1/4)x(tanx)^4-(1/12)(tanx)^3+(1/2)x(tanx)^2-(1/4)(tanx-x)+C